You have found the following ages (in years) of all 6 zebras at your local zoo: $ 17,\enspace 8,\enspace 6,\enspace 3,\enspace 18,\enspace 20$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{17 + 8 + 6 + 3 + 18 + 20}{{6}} = {12\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $17$ years $5$ years $25$ years $^2$ $8$ years $-4$ years $16$ years $^2$ $6$ years $-6$ years $36$ years $^2$ $3$ years $-9$ years $81$ years $^2$ $18$ years $6$ years $36$ years $^2$ $20$ years $8$ years $64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{25} + {16} + {36} + {81} + {36} + {64}} {{6}} $ $ {\sigma^2} = \dfrac{{258}}{{6}} = {43\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{43\text{ years}^2}} = {6.6\text{ years}} $ The average zebra at the zoo is 12 years old. There is a standard deviation of 6.6 years.